Complete Question:
A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.
Answer:
13 mol/L
Explanation:
The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:
M = n/V
The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:
n = 134/207.319
n = 0.646 mol
So, for a volume of 50 mL (0.05 L), the concentration is:
M = 0.646/0.05
M = 12.92 mol/L
Rounded to 2 significant digits, M = 13 mol/L
Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
I think D?? I apologize if not-check in other answers to be sure ^^
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Salts that are from strong bases and strong acids do not hydrolyze. Salts that are from strong bases and weak acids do hydrolyze, which gives it a pH greater than 7. Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7.
